Problem: $ D = \left[\begin{array}{rr}-2 & 0 \\ 2 & 3 \\ 1 & -2\end{array}\right]$ $ A = \left[\begin{array}{rr}1 & 2 \\ 2 & 1\end{array}\right]$ What is $ D A$ ?
Explanation: Because $ D$ has dimensions $(3\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ D A = \left[\begin{array}{rr}{-2} & {0} \\ {2} & {3} \\ \color{gray}{1} & \color{gray}{-2}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{2} \\ {2} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{0}\cdot{2} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{0}\cdot{2} & ? \\ {2}\cdot{1}+{3}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{1}+{0}\cdot{2} & {-2}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{1} \\ {2}\cdot{1}+{3}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{1}+{0}\cdot{2} & {-2}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{1} \\ {2}\cdot{1}+{3}\cdot{2} & {2}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{1} \\ \color{gray}{1}\cdot{1}+\color{gray}{-2}\cdot{2} & \color{gray}{1}\cdot\color{#DF0030}{2}+\color{gray}{-2}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-2 & -4 \\ 8 & 7 \\ -3 & 0\end{array}\right] $